I have this question:

A student takes a test of multiple option where each question has $5$ possible answers. If the student knows the correct answer, he/she selects it; in opposite case, he/she selects one randomly from the possibles $5$. Suppose that the student knows the answer of the $70\%$ of the questions.

-Which is the probability that from one given question the student gives the correct answer?

-If the student gets the correct answer to a question which is the probability that he/she knew that answer?

From the first question I thought that the answer could be $$\frac{7}{10}\cdot P(\text{correct}\mid \text{student doesn't know})=\frac{7}{10}\cdot\frac{\left(\frac{1}{5}\right)}{\left(\frac{3}{10}\right)}=\frac{7}{15}$$

But I don't know if it is correct. Is it? For the second question I don't know how it could be. Could anyone explain me how to tackle them?

  • 2
    $\begingroup$ How could this be right? There's a 70% chance the student knows the answer. How could the probability of his answering a question correctly be less than that? (7/15 < 7/10). You need to add, not multiply. (But you need to add the right thing, not what you have above.) For the second part, you need to use Bayes' rule. $\endgroup$ – saulspatz Jan 30 '18 at 18:12
  • $\begingroup$ Well, that's the reason because I wanted to ask, because I was doubting that it could be wrong, but I didn't know how it could be correct. $\endgroup$ – aavillanueva Jan 30 '18 at 18:16
  • $\begingroup$ Fair enough, but I think you should have put that in your question. "I know this can't be right because ... but I can't see what I'm doing wrong." This isn't meant as a criticism. I just mean that the more precisely you describe your problem the better help we'll be able to give you. $\endgroup$ – saulspatz Jan 30 '18 at 18:30

$\begin {array}{} &\text {Student Knows}&\text {Student Guesses}&\text {Totals}\\ \text {Correct}& 0.70 &0.30\cdot 0.20 = 0.06&0.76\\ \text {Incorrect}& 0 &0.30\cdot 0.80 = 0.24&0.24\end {array}$

$P(\text {Knows}|\text { Correct}) = \frac {0.70}{0.76}$

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No, your answer and approach for the first question is not correct.

Think about it like this:

There are two ways for the student to get the correct answer: either the student knows the answer, or the student does not know, but guessed right. These are mutually exclusive events, and so you should add the probabilities for each of those events to get the probability of getting the correct answer.

So, if we use $C$ for the evnt of the student picking the correct answer, and $K$ for the event of the student knowing the answer, we get:

$$P(C) = P(C \cap K)+ P(C \cap K^C)$$


$$P(C \cap K) = P(C|K)\cdot P(K)$$

$$P(C \cap K^C) = P(C|K^C)\cdot P(K^C)$$

And we are given that:


and thus:

$$P(K^C) = 1-0.7=0.3$$






$$P(C)= P(C|K)\cdot P(K)+ P(C|K^C)\cdot P(K^C)=1 \cdot 0.7 + 0.2 \cdot 0.3 = 0.76$$

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The formula (7/10)(P(correct|student doesn't know)) ignores the possibility that the student knows the answer, and your calculations from there incorrectly evaluates the expression P(correct|student doesn't know).

One strategy is to pick a number of questions such that all the numbers end up being integers. The denominators involved are 5 (five possible answers) and 10 (70% chance knows the answer). So if we multiply those together, we get 50. Draw out a grid with two rows and two columns corresponding to "knows answers"/"doesn't know" and "gets answer right"/"gets answer wrong", and fill out the boxes. Then for each probability asked for, figure out what the population set is, and what the "success" set is, and find the ratio of the two.

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